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1
becuase the same path followed so displacement is zero and distance is 5 + 5 = 10
Answer:
Answer :
A body is taken 32 km above the surface of the earth
Radius of earth = 6400 km
Percentage decrease in weight of the body = ?
\quad ━━━━━━━━━━━━━━━━━━
\begin{gathered}\sf :\implies g' = g\bigg\lgroup 1 + \dfrac{h}{R}\bigg\rgroup{}^{-2}\;\; -eq(1)\\\end{gathered}
:⟹g
′
=g
⎩
⎪
⎪
⎪
⎧
1+
R
h
⎭
⎪
⎪
⎪
⎫
−2
−eq(1)
\qquad\quad\dag\:\small\sf h < < R†h<<R
\qquad\quad\dag\:\small\sf \dfrac{h}{R} < < 1†
R
h
<<1
\begin{gathered}\sf :\implies (1+x)^{-2} = 1 - 2x \;\; x < < 1\\\end{gathered}
:⟹(1+x)
−2
=1−2xx<<1
\sf :\implies \bigg\lgroup 1 + \dfrac{h}{R}\bigg\rgroup{}^{-2} = 1 - \dfrac{2h}{R}:⟹
⎩
⎪
⎪
⎪
⎧
1+
R
h
⎭
⎪
⎪
⎪
⎫
−2
=1−
R
2h
\begin{gathered}\sf :\implies g' = g\bigg\lgroup 1 - \dfrac{2h}{R}\bigg\rgroup\;\; -eq(2)\\\end{gathered}
:⟹g
′
=g
⎩
⎪
⎪
⎪
⎧
1−
R
2h
⎭
⎪
⎪
⎪
⎫
−eq(2)
So then the % decrease in the value of acceleration due to gravity is gonna be,
\begin{gathered}\sf :\implies \delta g = \dfrac{g' - g}{g} \times 100\\\end{gathered}
:⟹δg=
g
g
′
−g
×100
\begin{gathered}\sf :\implies \delta g = \dfrac{g\bigg\lgroup 1 - \dfrac{2h}{R}\bigg\rgroup - g}{g} \times 100\\\end{gathered}
:⟹δg=
g
g
⎩
⎪
⎪
⎪
⎧
1−
R
2h
⎭
⎪
⎪
⎪
⎫
−g
×100
\begin{gathered}\sf :\implies \delta g = \dfrac{g\bigg\{ \bigg\lgroup 1 - \dfrac{2h}{R}\bigg\rgroup - 1 \bigg\} }{g} \times 100\\\end{gathered}
:⟹δg=
g
g{
⎩
⎪
⎪
⎪
⎧
1−
R
2h
⎭
⎪
⎪
⎪
⎫
−1}
×100
\begin{gathered}\sf :\implies \delta g = \bigg\{ 1 - \dfrac{2h}{R} - 1 \bigg\} 100\\\end{gathered}
:⟹δg={1−
R
2h
−1}100
\sf :\implies \delta g = - \dfrac{200h}{R} \;\; -eq(3):⟹δg=−
R
200h
−eq(3)
Finally, the % change in weight will be given by,
\begin{gathered}\sf :\implies \delta W = \delta m + \delta g\\\end{gathered}
:⟹δW=δm+δg
\begin{gathered}\sf :\implies \delta W = - \dfrac{200h}{R}\\\end{gathered}
:⟹δW=−
R
200h
\displaystyle \underline{\bigstar\:\textsf{According to the Question :}}
★According to the Question :
\sf\dashrightarrow \delta W = \dfrac{200\times 32}{6400}⇢δW=
6400
200×32
\underline{\boxed{\pink{\mathfrak {\delta w = -1 \%}}}}
δw=−1%